Program Analysis and Verification 0368-4479 Noam Rinetzky Lecture 3: Axiomatic Semantics 1 Slides...

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Program Analysis and Verification

0368-4479

Noam Rinetzky

Lecture 3: Axiomatic Semantics

1

Slides credit: Roman Manevich, Mooly Sagiv, Eran Yahav

Axiomatic Semantics

C.A.R. HoareRobert W. Floyd

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Edsger W. Dijkstra

Axiomatic Semantics

C.A.R. HoareRobert W. Floyd

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Edsger W. Dijkstra

BTW, what do all these people have in common?

Axiomatic Semantics

C.A.R. HoareRobert W. Floyd

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Edsger W. Dijkstra

1972 1978 1980For having a clear influence on methodologies for the creation of efficient and reliable software, and for helping to found the following important subfields of computer science: the theory of parsing, the semantics of programming languages, automatic program verification, automatic program synthesis, and analysis of algorithms.

For fundamental contributions to programming as a high, intellectual challenge; for eloquent insistence and practical demonstration that programs should be composed correctly, not just debugged into correctness; for illuminating perception of problems at the foundations of program design.

For his fundamental contributions to the definition and design of programming languages.

http://amturing.acm.org/

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Proving program correctness

• Why prove correctness?• What is correctness?• How?– Reasoning at the operational semantics level• Tedious• Error prone

– Formal reasoning using “axiomatic” semantics • Syntactic technique (“game of tokens”)• Mechanically checkable

– Sometimes automatically derivable

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A simple imperative language: While

Abstract syntax:a ::= n | x | a1 + a2 | a1 a2 | a1 – a2

b ::= true | false| a1 = a2 | a1 a2 | b | b1 b2

S ::= x := a | skip | S1; S2

| if b then S1 else S2

| while b do S

7

Program correctness concepts

• Property = a certain relationship between initial state and final state

• Partial correctness = properties that holdif program terminates

• Termination = program always terminates– i.e., for every input state

partial correctness + termination = total correctness

Other correctness concepts exist: resource usage, linearizability, …

Mostly focus in this course

Other notions of properties exist

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Factorial example

• Sfac , s s’ implies s’ y = (s x)!

Sfac y := 1; while (x=1) do (y := y*x; x := x–1)

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Factorial example

• Sfac , s s’ implies s’ y = (s x)!

• Factorial partial correctness property = – if the statement terminates then the final value of y will be the factorial of the initial value of x• What if s x < 0?

Sfac y := 1; while (x=1) do (y := y*x; x := x–1)

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Natural semantics for While

x := a, s s[x Aas][assns]

skip, s s[skipns]

S1, s s’, S2, s’ s’’S1; S2, s s’’

[compns]

S1, s s’ if b then S1 else S2, s s’

if B b s = tt[ifttns]

S2, s s’ if b then S1 else S2, s s’

if B b s = ff[ifffns]

while b do S, s s if B b s = ff[whileffns]

S, s s’, while b do S, s’ s’’while b do S, s s’’

if B b s = tt[whilettns]

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Staged proof

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First stage

Second stage

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while (x=1) do (y := y*x; x := x–1), s s’

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Third stage

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How easy was that?

• Proof is very laborious– Need to connect all transitions and argues about

relationships between their states– Reason: too closely connected to semantics of

programming language

• Is the proof correct?

• How did we know to find this proof?– Is there a methodology?

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Axiomatic verification approach

• What do we need in order to prove that the program does what it supposed to do?

• Specify the required behavior

• Compare the behavior with the one obtained by the operational semantics

• Develop a proof system for showing that the program satisfies a requirement

• Mechanically use the proof system to show correctness

• The meaning of a program is a set of verification rules

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Axiomatic Verification: Spec

• Sfac , s s’ implies s’ y = (s x)!

• {x = N} Sfac {y = N!} – {Pre-condition (s)} Command (Sfac) {post-state(s’)} – Not {true} Sfac {y = x!}

Sfac y := 1; while (x=1) do (y := y*x; x := x–1)

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Partial vs. Total Correctness

• Sfac , s s’ implies s’ y = (s x)!

• {x = N} Sfac {y = N!} – {Pre-condition (s)} Command (Sfac) {post-state(s’)}

– Not {true} Sfac {y = x!}

• [x = N] Sfac [y = N!]

Sfac y := 1; while (x=1) do (y := y*x; x := x–1)

Hoare Triples

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Verification: Assertion-Based [Floyd, ‘67]

• Assertion: invariant at specific program point – E.g., assert(e)

• use assertions as foundation for static correctness proofs

• specify assertions at every program point• correctness reduced to reasoning about

individual statements

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Annotated Flow ProgramsReduction: Program verification is reduced to claims about the subject of discourse

Straight line code: claims are determined

“by construction”

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Annotated Flow ProgramsReduction: Program verification is reduced to claims about the subject of discourse

Straight line code: claims are determined “by

construction”

Cut points

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Assertion-Based Verification [Floyd, ‘67]

• Assertion: invariant at specific program point – E.g., assert(e)

• Proof reduced to logical claims– Considering the effect of statements– But, not reusable

• Challenge: Finding invariants at cut points in loops

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Floyd-Hoare Logic 1969

• Use Floyd’s ideas to define axiomatic semantics– Structured programming• No gotos• Modular (reusable claims)

– Hoare triples• {P} C {Q}• [P] C [Q] (often <P> C <Q>)

– Define the programming language semantics as a proof system

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Assertions, a.k.a Hoare triples

• P and Q are state predicates– Example: x>0

• If P holds in the initial state, andif execution of C terminates on that state,then Q will hold in the state in which C halts

• C is not required to always terminate {true} while true do skip {false}

{ P } C { Q }precondition postcondition

statementa.k.a command

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Total correctness assertions

• If P holds in the initial state,execution of C must terminate on that state,and Q will hold in the state in which C halts

[ P ] C [ Q ]

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Factorial example

{ ? } y := 1; while (x=1) do (y := y*x; x := x–1)

{ ? }

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First attempt

{ x>0 } y := 1; while (x=1) do (y := y*x; x := x–1)

{ y=x! }

Holds only for value of x at state after execution finishes

We need a way to “remember” value of x before execution

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Fixed assertion

{ x=n } y := 1; while (x=1) do (y := y*x; x := x–1)

{ y=n! n>0 }

A logical variable, must not appear in statement - immutable

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The proof outline

{ x=n } y := 1;{ x>0 y*x!=n! nx } while (x=1) do { x-1>0 (y*x)*(x-1)!=n! n(x-1) } y := y*x; { x-1>0 y*(x-1)!=n! n(x-1) } x := x–1{ y*x!=n! n>0 x=1 }

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Factorial example

• Factorial partial correctness property = if the statement terminates then the final value of y will be the factorial of the initial value of x– What if s x < 0?

• Formally, using natural semantics:Sfac , s s’ implies s’ y = (s x)!

Sfac y := 1; while (x=1) do (y := y*x; x := x–1)

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Staged proof

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Stages

y := 1; while (x=1) do (y := y*x; x := x–1)s s’

s’ y = (s x)! s x > 0

while (x=1) do (y := y*x; x := x–1)

y := y*x; x := x–1s s’’

s y (s x)! = s’’ y (s’’ x)! s x > 0

s s’’

s y (s x)! = s’’ y (s’’ x)! s’’x = 1 s x > 0

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Inductive proof over iterations

while (x=1) do (y := y*x; x := x–1)

(y := y*x; x := x–1)

while (x=1) do (y := y*x; x := x–1)

s s’’s y (s x)! = s’’ y (s’’ x)! s’’x = 1 s x > 0

s s’s’ s’’

s’ y (s’ x)! = s’’ y (s’’ x)! s’’x = 1 s’ x > 0

s y (s x)! = s’ y (s’ x)! s x > 0

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Assertions, a.k.a Hoare triples

• P and Q are state predicates– Example: x>0

• If P holds in the initial state, andif execution of C terminates on that state,then Q will hold in the state in which C halts

• C is not required to always terminate {true} while true do skip {false}

{ P } C { Q }

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Total correctness assertions

• If P holds in the initial state,execution of C must terminate on that state,and Q will hold in the state in which C halts

[ P ] C [ Q ]

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Factorial assertion

{ x=n } y := 1; while (x=1) do (y := y*x; x := x–1)

{ y=n! n>0 }

A logical variable, must not appear in statement - immutable

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Factorial partial correctness proof

{ x=n } y := 1;{ x>0 y*x!=n! nx } while (x=1) do { x-1>0 (y*x)*(x-1)!=n! n(x-1) } y := y*x; { x-1>0 y*(x-1)!=n! n(x-1) } x := x–1{ y*x!=n! n>0 x=1 }

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P

Formalizing partial correctness

• s P– P holds in state s

• – program states – undefined

Sns C s =

s’ if C, s s’ else

s

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Formalizing partial correctness

• s P– P holds in state s

• – program states – undefined

• { P } C { Q }– s, s’ . (sP C, ss’) s’Q

alternatively– s . (sP SnsC s) SnsC Q– Convention: P for all P

s . sP SnsC s Q

P C(P)

Q

ss’C

Why did we choose natural semantics?

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Formalizing partial correctness

• s P– P holds in state s

• – program states – undefined

• { P } C { Q }– s, s’ . (sP C, s*s’) s’Q

alternatively– s . (sP SsosC s) SsosC Q– Convention: P for all P

s . sP SsosC s Q

P C(P)

Q

ss’C

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How do we express predicates?

• Extensional approach– Abstract mathematical functions

P : State T

• Intensional approach– Via language of formulae

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An assertion language

• Bexp is not expressive enough to express predicates needed for many proofs– Extend Bexp

• Allow quantifications– z. … – z. … • z. z = kn

• Import well known mathematical concepts– n! n (n-1) 2 1

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An assertion language

a ::= n | x | a1 + a2 | a1 a2 | a1 – a2

A ::= true | false| a1 = a2 | a1 a2 | A | A1 A2 | A1 A2

| A1 A2 | z. A | z. A

Either a program variables or a logical variable

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First Order Logic Reminder

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Free/bound variables

• A variable is said to be bound in a formula when it occurs in the scope of a quantifier. Otherwise it is said to be free– i. k=im– (i+10077)i. j+1=i+3)

• FV(A) the free variables of A• Defined inductively on the abstract syntax tree

of A

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Free variables

FV(n) {}FV(x) {x}FV(a1+a2) FV(a1a2) FV(a1-a2) FV(a1) FV(a2)FV(true) FV(false) {}FV(a1=a2) FV(a1a2) FV(a1) FV(a2)FV(A) FV(A)FV(A1 A2) FV(A1 A2) FV(A1 A2)FV(z. A) FV(z. A) FV(A) \ {z}

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Substitution

• An expression t is pure (a term) if it does not contain quantifiers

• A[t/z] denotes the assertion A’ which is the same as A, except that all instances of the free variable z are replaced by t

• A i. k=imA[5/k] = A[5/i] =

What if t is not pure?

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Calculating substitutions

n[t/z] = nx[t/z] = xx[t/x] = t

(a1 + a2)[t/z] = a1[t/z] + a2[t/z](a1 a2)[t/z] = a1[t/z] a2[t/z](a1 - a2)[t/z] = a1[t/z] - a2[t/z]

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Calculating substitutionstrue[t/x] = truefalse[t/x] = false(a1 = a2)[t/z] = a1[t/z] = a2[t/z] (a1 a2)[t/z] = a1[t/z] a2[t/z] (A)[t/z] = (A[t/z])(A1 A2)[t/z]= A1[t/z] A2[t/z](A1 A2)[t/z] = A1[t/z] A2[t/z] (A1 A2)[t/z] = A1[t/z] A2[t/z]

(z. A)[t/z] = z. A(z. A)[t/y] = z. A[t/y]( z. A)[t/z] = z. A( z. A)[t/y] = z. A[t/y]

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Proof Rules

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Axiomatic semantics for While { P[a/x] } x := a { P }[assp]

{ P } skip { P }[skipp]

{ P } S1 { Q }, { Q } S2 { R } { P } S1; S2 { R }[compp]

{ b P } S1 { Q }, { b P } S2 { Q } { P } if b then S1 else S2 { Q }[ifp]

{ b P } S { P } { P } while b do S {b P }

[whilep]

{ P’ } S { Q’ } { P } S { Q }

[consp] if PP’ and Q’Q

Notice similarity to natural semantics rules

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Assignment rule

• A “backwards” rule• x := a always finishes• Why is this true?– Recall operational semantics:

• Example: {y*z<9} x:=y*z {x<9}What about {y*z<9w=5} x:=y*z {w=5}?

x := a, s s[xAas][assns]

s[xAas] P

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skip rule

skip, s s[skipns]

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Composition rule

• Holds when S1 terminates in every state where P holds and then Q holdsand S2 terminates in every state where Q holds and then R holds

S1, s s’, S2, s’ s’’S1; S2, s s’’ [compns]

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Condition rule

S1, s s’ if b then S1 else S2, s s’

if B b s = tt[ifttns]

S2, s s’ if b then S1 else S2, s s’

if B b s = ff[ifffns]

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Loop rule

• Here P is called an invariant for the loop– Holds before and after each loop iteration– Finding loop invariants – most challenging part of proofs

• When loop finishes, b is false

while b do S, s s if B b s = ff[whileffns]

S, s s’, while b do S, s’ s’’while b do S, s s’’

if B b s = tt[whilettns]

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Rule of consequence

• Allows strengthening the precondition and weakening the postcondition

• The only rule that is not sensitive to the form of the statement

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Rule of consequence

• Why do we need it?• Allows the following

{y*z<9} x:=y*z {x<9} {y*z<9w=5} x:=y*z {x<10}

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Axiomatic semantics for While { P[a/x] } x := a { P }[assp]

{ P } skip { P }[skipp]

{ P } S1 { Q }, { Q } S2 { R } { P } S1; S2 { R }[compp]

{ b P } S1 { Q }, { b P } S2 { Q } { P } if b then S1 else S2 { Q }[ifp]

{ b P } S { P } { P } while b do S {b P }

[whilep]

{ P’ } S { Q’ } { P } S { Q }

[consp] if PP’ and Q’Q

Inference rule for every composed statement

Axiom for every primitive statement

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Inference trees

• Similar to derivation trees of natural semantics• Leaves are …• Internal nodes correspond to …• Inference tree is called– Simple if tree is only an axiom– Composite otherwise

• Similar to derivation trees of natural semantics– Reasoning about immediate constituent

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Factorial proof

W = while (x1) do (y:=y*x; x:=x–1)

INV = x > 0 (y x! = n! n x)

{ INV[x-1/x][y*x/y] } y:=y*x; x:=x–1 {INV}

{ INV[x-1/x] } x:=x-1 {INV}

{ INV } W {x=1 INV }{ INV[1/y] } y:=1 { INV }

{ INV[x-1/x][y*x/y] } y:=y*x { INV[x-1/x] }

{x1 INV } y:=y*x; x:=x–1 { INV }

[comp]

[cons]

[while]

[cons]{ INV } W { y=n! n>0 }{ x=n } y:=1 { INV }

[cons]

{ x=n } while (x1) do (y:=y*x; x:=x–1) { y=n! n>0 }

[comp]

Goal: { x=n } y:=1; while (x1) do (y:=y*x; x:=x–1) { y=n! n>0 }

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Factorial proof

W = while (x1) do (y:=y*x; x:=x–1)

INV = x > 0 (y x! = n! n x)

{ INV[x-1/x][y*x/y] } y:=y*x; x:=x–1 {INV}

{ INV[x-1/x] } x:=x-1 {INV}

{ INV } W {x=1 INV }{ INV[1/y] } y:=1 { INV }

{ INV[x-1/x][y*x/y] } y:=y*x { INV[x-1/x] }

{x1 INV } y:=y*x; x:=x–1 { INV }

[comp]

[cons]

[while]

[cons]{ INV } W { y=n! n>0 }{ x=n } y:=1 { INV }

[cons]

{ x=n } while (x1) do (y:=y*x; x:=x–1) { y=n! n>0 }

[comp]

Goal: { x=n } y:=1; while (x1) do (y:=y*x; x:=x–1) { y=n! n>0 }

{ b P } S { P } { P } while b do S {b P }

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Factorial proof

W = while (x1) do (y:=y*x; x:=x–1)

INV = x > 0 (y x! = n! n x)

{ INV[x-1/x][y*x/y] } y:=y*x; x:=x–1 {INV}

{ INV[x-1/x] } x:=x-1 {INV}

{ INV } W {x=1 INV }{ INV[1/y] } y:=1 { INV }

{ INV[x-1/x][y*x/y] } y:=y*x { INV[x-1/x] }

{x1 INV } y:=y*x; x:=x–1 { INV }

[comp]

[cons]

[while]

[cons]{ INV } W { y=n! n>0 }{ x=n } y:=1 { INV }

[cons]

{ x=n } while (x1) do (y:=y*x; x:=x–1) { y=n! n>0 }

[comp]

Goal: { x=n } y:=1; while (x1) do (y:=y*x; x:=x–1) { y=n! n>0 }

{ P’ } S { Q’ } { P } S { Q } if PP’ and Q’Q

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Provability

• We say that an assertion { P } C { Q } is provable if there exists an inference tree– Written as p { P } C { Q }

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Annotated programs

• A streamlined version of inference trees– Inline inference trees into programs– A kind of “proof carrying code”– Going from annotated program to proof tree is a

linear time translation

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Annotating composition

• We can inline inference trees into programs• Using proof equivalence of S1; (S2; S3) and (S1; S2); S3

instead writing deep trees, e.g.,

{P} (S1; S2); (S3 ; S4) {Q}{P} (S1; S2) {P’’} {P’’} (S3 ; S4) {Q}

{P} S1 {P’} {P’} S2 {P’’} {P’’} S3 {P’’’} {P’’’} S4 {P’’}

• We can annotate a composition S1; S2;…; Sn by{P1} S1 {P2} S2 … {Pn-1} Sn-1 {Pn}

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Annotating conditions

{ P }if b then { b P } S1

else S2

{ Q }

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Annotating conditions

{ P }if b then

{ b P }S1

{ Q1 }else

S2

{ Q2 }{ Q }

Usually Q is the result of using the consequence rule, so a more explicit annotation is

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Annotating loops

{ P }while b do { b P } S{b P }

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Annotating loops

{ P }while b do { b P } S { P’ }{b P } { Q }

P’ implies P

b P implies Q

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Annotated factorial program{ x=n } y := 1;{ x>0 y*x!=n! nx } while (x=1) do { x-1>0 (y*x)*(x-1)!=n! n(x-1) } y := y*x; { x-1>0 y*(x-1)!=n! n(x-1) } x := x–1{ y*x!=n! n>0 }

• Contrast with proof via natural semantics

• Where did the inductive argument over loop iterations go?

73

Properties of the semanticsEquivalence– What is the analog of program

equivalence in axiomatic verification?

Soundness– Can we prove incorrect properties?

Completeness– Is there something we can’t

prove?

74

Provability

• We say that an assertion { P } C { Q } is provable if there exists an inference tree– Written as p { P } C { Q }– Are inference trees unique?

{true} x:=1; x:=x+5 {x0}• Proofs of properties of axiomatic semantics use

induction on the shape of the inference tree– Example: prove p { P } C { true } for any P and C

75

Provable equivalence

• We say that C1 and C2 are provably equivalent if for all P and Qp { P } C1 { Q } if and only if p { P } C2 { Q }

• Examples:– S; skip and S– S1; (S2; S3) and (S1; S2); S3

76

Valid assertions

• We say that { P } C { Q } is valid if for all states s, if sP and C, ss’ then s’Q

• Denoted by p { P } C { Q }

P C(P)

Q

ss’C

77

Logical implication and equivalence

• We write A B if for all states sif s A then s B– {s | s A } {s | s B }– For every predicate A: false A true

• We write A B if A B and B A– false 5=7

• In writing Hoare-style proofs, we will often replace a predicate A with A’ such that A A’and A’ is “simpler”

78

Soundness and completeness

• The inference system is sound:– p { P } C { Q } implies p { P } C { Q }

• The inference system is complete:– p { P } C { Q } implies p { P } C { Q }

79

Hoare logic is sound and (relatively) complete

• Soundness: p { P } C { Q } implies p { P } C { Q }

• (Relative) completeness: p { P } C { Q } implies p { P } C { Q }

– Provided we can prove any implication RR’

80

Hoare logic is sound and (relatively) complete

• Soundness: p { P } C { Q } implies p { P } C { Q }

• (Relative) completeness: p { P } C { Q } implies p { P } C { Q }

– Provided we can prove any implication RR’• FYI, nobody tells us how to find a proof …

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Is there an Algorithm?{ x=n } y := 1;{ x>0 y*x!=n! nx } while (x=1) do { x-1>0 (y*x)*(x-1)!=n! n(x-1) } y := y*x; { x-1>0 y*(x-1)!=n! n(x-1) } x := x–1{ y*x!=n! n>0 }

Annotated programs provides a compact representation of inference trees

82

?

83

Predicate Transformers

84

Weakest liberal precondition

• A backward-going predicate transformer• The weakest liberal precondition for Q is

s wlp(C, Q)if and only if for all states s’if C, ss’ then s’ Q

Propositions:1. p { wlp(C, Q) } C { Q }

2. If p { P } C { Q } then P wlp(C, Q)

85

Strongest postcondition

• A forward-going predicate transformer• The strongest postcondition for P is

s’ sp(P, C)if and only if there exists s such thatif C, ss’ and s P

1. p { P } C { sp(P, C) }

2. If p { P } C { Q } then sp(P, C) Q

86

Predicate transformer semantics• wlp and sp can be seen functions that transform

predicates to other predicates– wlpC : Predicate Predicate

{ P } C { Q } if and only if wlpC Q = P– spC : Predicate Predicate

{ P } C { Q } if and only if spC P = Q

87

Hoare logic is (relatively) complete• Proving

p { P } C { Q } implies p { P } C { Q }is the same as provingp { wlp(C, Q) } C { Q }

• Suppose that p { P } C { Q }then (from proposition 2) P { wlp(C, Q) }

{ P } S { Q } { wlp(C, Q) } S { Q }

[consp]

88

Calculating wlp

1. wlp(skip, Q) = Q2. wlp(x := a, Q) = Q[a/x]3. wlp(S1; S2, Q) = wlp(S1, wlp(S2, Q))

4. wlp(if b then S1 else S2, Q) =(b wlp(S1, Q)) (b wlp(S2,

Q))5. wlp(while b do S, Q) = … ?

hard to capture

89

Calculating wlp of a loop

wlp(while b do S, Q) =

Idea: we know the following statements are semantically equivalentwhile b do Sif b do (S; while b do S) else skip

Let’s try to substitute and calculate on

wlp(if b do (S; while b do S) else skip, Q) =

(b wlp(S; while b do S, Q)) (b wlp(skip, Q)) =

(b wlp(S, wlp(while b do S, Q))) (b Q)

LoopInv = (b wlp(S, LoopInv)) (b Q) We have a recurrence

The loop invariant

90

Prove the following triple

• LoopInv = (b wlp(S, LoopInv)) (b Q)• Let’s substitute LoopInv with timer0• Show that timer0 is equal to

(timer>0 wlp(timer:=timer-1, timer0)) (timer0 timer=0)= (timer>0 (timer0)[timer-1/timer]) (timer0 timer=0)= (timer>0 timer-10) (timer0 timer=0)= timer>0 timer=0= timer0

{ timer 0 }while (timer > 0) do

timer := timer – 1{ timer = 0 }

91

Issues with wlp-based proofs

• Requires backwards reasoning – not very intuitive

• Backward reasoning is non-deterministic – causes problems when While is extended with dynamically allocated heaps (aliasing)

• Also, a few more rules will be helpful

92

Conjunction rule

• Not necessary (for completeness) but practically useful

• Starting point of extending Hoare logic to handle parallelism

• Related to Cartesian abstraction– Will point this out when we learn it

93

Structural Rules{ P } C { Q } { P’ } C { Q’ }

{ P P’ } C {Q Q’ }[disjp]

{ P } C { Q } { v. P } C { v. Q }[existp] vFV(C

)

{ P } C { Q } {v. P } C {v. Q }[univp] vFV(C)

{ F } C { F } Mod(C) FV(F)={}[Invp]• Mod(C) = set of variables assigned to in sub-statements of C• FV(F) = free variables of F

94

Invariance + Conjunction = Constancy

• Mod(C) = set of variables assigned to in sub-statements of C• FV(F) = free variables of F

{ P } C { Q } { F P } C { F Q }[constancyp] Mod(C) FV(F)={}

95

Floyd’s strongest postcondition rule

• Example{ z=x } x:=x+1 { ?v. x=v+1 z=v }

• This rule is often considered problematic because it introduces a quantifier – needs to be eliminated further on

• We will now see a variant of this rule

{ P } x := a { v. x=a[v/x] P[v/x] } where v is a fresh variable

[assFloyd]

96

“Small” assignment axiom

• Examples:{x=n} x:=5*y {x=5*y}{x=n} x:=x+1 {x=n+1}

{x=y} x:=y+1 {x=y+1}{x=n} x:=y+1 {x=y+1}[existp] {n. x=n} x:=y+1 {n. x=y+1} therefore {true} x:=y+1 {x=y+1} [constancyp] {z=9} x:=y+1 {z=9 x=y+1}

{ x=v } x:=a { x=a[v/x] }where vFV(a)

[assfloyd]

First evaluate ain the precondition state(as a may access x)

Then assign the resulting value to x

Create an explicit Skolem variable in precondition

97

Buggy sum program{ y0 }x := 0{ y0 x=0 }res := 0{ y0 x=0 res=0 }Inv = { y0 res=Sum(0, x) } = { y0 res=m x=n m=Sum(0, n) } while (xy) do

{ y0 res=m x=n m=Sum(0, n) xy ny } x := x+1{ y0 res=m x=n+1 m=Sum(0, n) ny}res := res+x { y0 res=m+x x=n+1 m=Sum(0, n) ny}{ y0 res-x=Sum(0, x-1) ny}{ y0 res=Sum(0, x) }

{ y0 res=Sum(0, x) x>y } {res = Sum(0, y) }

98

Sum program• Define Sum(0, n) = 0+1+…+n

{ y0 }x := 0{ y0 x=0 }res := 0{ y0 x=0 res=0 }Inv = { y0 res=Sum(0, x) xy } { y0 res=m x=n ny m=Sum(0, n) } while (x<y) do

{ y0 res=m x=n m=Sum(0, n) x<y n<y }res := res+x{ y0 res=m+x x=n m=Sum(0, n) n<y }x := x+1 { y0 res=m+x x=n+1 m=Sum(0, n) n<y }{ y0 res-x=Sum(0, x-1) x-1<y }{ y0 res=Sum(0, x) }

{ y0 res=Sum(0, x) xy xy }{ y0 res=Sum(0, y) x=y }{ res = Sum(0, y) }

{ x=Sum(0, n) } { y=n+1 }

{ x+y=Sum(0, n+1) }

Background axiom

99

Floyd’s strongest postcondition rule

• Example{ z=x } x:=x+1 { ?v. x=v+1 z=v }

• This rule is often considered problematic because it introduces a quantifier – needs to be eliminated further on

• We will now see a variant of this rule

{ P } x := a { v. x=a[v/x] P[v/x] } where v is a fresh variable

[assFloyd]

100

Floyd’s strongest postcondition rule

• Example{ z=x } x:=x+1 { v. x=v+1 z=v }

• This rule is often considered problematic because it introduces a quantifier – needs to be eliminated further on

• We will now see a variant of this rule

{ P } x := a { v. x=a[v/x] P[v/x] } where v is a fresh variable

[assFloyd]

101

“Small” assignment axiom

• Examples:{x=n} x:=5*y {x=5*y}{x=n} x:=x+1 {x=n+1}

{x=n} x:=y+1 {x=y+1}[existp] {n. x=n} x:=y+1 {n. x=y+1} therefore {true} x:=y+1 {x=y+1} [constancyp] {z=9} x:=y+1 {z=9 x=y+1}

{ x=v } x:=a { x=a[v/x] }where vFV(a)

[assfloyd]

First evaluate ain the precondition state(as a may access x)

Then assign the resulting value to x

Create an explicit Skolem variable in precondition

102

“Small” assignment axiom

• Examples:{x=n} x:=5*y {x=5*y}{x=n} x:=x+1 {x=n+1}

{x=n} x:=y+1 {x=y+1}[existp] {n. x=n} x:=y+1 {n. x=y+1} therefore {true} x:=y+1 {x=y+1} [constancyp] {z=9} x:=y+1 {z=9 x=y+1}

{ x=v } x:=a { x=a[v/x] }where vFV(a)

[assfloyd]

103

“Small” assignment axiom

• Examples:{x=n} x:=5*y {x=5*y}{x=n} x:=x+1 {x=n+1}

{x=n} x:=y+1 {x=y+1}[existp] {n. x=n} x:=y+1 {n. x=y+1} therefore {true} x:=y+1 {x=y+1} [constancyp] {z=9} x:=y+1 {z=9 x=y+1}

{ x=v } x:=a { x=a[v/x] }where vFV(a)

[assfloyd]

104

“Small” assignment axiom

• Examples:{x=n} x:=5*y {x=5*y}{x=n} x:=x+1 {x=n+1}

{x=n} x:=y+1 {x=y+1}[existp] {n. x=n} x:=y+1 {n. x=y+1} therefore {true} x:=y+1 {x=y+1} [constancyp] {z=9} x:=y+1 {z=9 x=y+1}

{ x=v } x:=a { x=a[v/x] }where vFV(a)

[assfloyd]

105

Buggy sum program{ y0 }x := 0{ y0 x=0 }res := 0{ y0 x=0 res=0 }Inv = { y0 res=Sum(0, x) } = { y0 res=m x=n m=Sum(0, n) } while (xy) do

{ y0 res=m x=n m=Sum(0, n) xy ny } x := x+1{ y0 res=m x=n+1 m=Sum(0, n) ny}res := res+x { y0 res=m+x x=n+1 m=Sum(0, n) ny}{ y0 res-x=Sum(0, x-1) ny}{ y0 res=Sum(0, x) }

{ y0 res=Sum(0, x) x>y } {res = Sum(0, y) }

106

Sum program• Define Sum(0, n) = 0+1+…+n{ y0 }x := 1{ y0 x=1 }res := 0{ y0 x=1 res=0 }Inv = { y0 res=Sum(0, x-1) xy+1 } { y0 res=m x=n ny+1 m=Sum(0, n-1) } while (xy) do

{ y0 res=m x=n m=Sum(0, n-1) x<y ny+1 }res := res+x{ y0 res=m+x x=n m=Sum(0, n-1) ny+1 }x := x+1 { y0 res=m+x x=n+1 m=Sum(0, n-1) ny+1 }{ y0 res-x=Sum(0, x-1) x-1<y+1 }{ y0 res=Sum(0, x-1) xy+1 } // axm-Sum

{ y0 res=Sum(0, x-1) xy+1 x>y }{ y0 res=Sum(0, x-1) x=y+1 }{ y0 res=Sum(0, y) }{ res = Sum(0, y) }

Background axiom

{ x=Sum(0, n) } { y=n+1 } { x+y=Sum(0, n+1) }

[axm-Sum]

107

Sum program• Define Sum(0, n) = 0+1+…+n

{ y0 }x := 1{ y0 x=0 }res := 0{ y0 x=0 res=0 }Inv = { y0 res=Sum(0, x-1) xy+1 }while (xy) do

{ y0 res=m x=n m=Sum(0, n-1) ny+1 x<y }res := res+x{ y0 res=m+x x=n m=Sum(0, n-1) ny+1 }{ y0 res=Sum(0, n) x=n ny+1 } // axm-Sum x := x+1{ y0 res=Sum(0, n) x=n+1 ny+1 }{ y0 res=Sum(0, x-1) xy+1 }

{ y0 res=Sum(0, x-1) xy+1 x>y }{ y0 res=Sum(0, x-1) x=y+1 }{ y0 res=Sum(0, y) }{ res = Sum(0, y) }

Background axiom

{ x=Sum(0, n) } { y=n+1 } { x+y=Sum(0, n+1) }

[axm-Sum]

108

Example 1: Absolute value program

{ }if x<0 then

x := -xelse

skip{ }

109

Absolute value program

{ x=v }if x<0 then { x=v x<0 }

x := -x { x=-v x>0 }else

{ x=v x0 }skip{ x=v x0 }

{ v<0 x=-v v0 x=v}{ x=|v| }

110

Example 2: Variable swap program

{ }t := xx := yy := t{ }

111

Variable swap program

{ x=a y=b }t := x{ x=a y=b t=a }x := y{ x=b y=b t=a }y := t{ x=b y=a t=a }{ x=b y=a } // cons

112

Example 3: Axiomatizing data types

• We added a new type of variables – array variables– Model array variable as a function y : Z Z

• We need the two following axioms:

S ::= x := a | x := y[a] | y[a] := x | skip | S1; S2

| if b then S1 else S2

| while b do S

{ y[xa](x) = a }

{ zx y[xa](z) = y(z) }

113

Array update rules (wp)

• Treat an array assignment y[a] := x as an update to the array function y– y := y[ax] meaning y’=v. v=a ? X : y(v)

S ::= x := a | x := y[a] | y[a] := x | skip | S1; S2

| if b then S1 else S2

| while b do S

[array-update] { P[y[ax]/y] } y[a] := x { P }

[array-load] { P[y(a)/x] } x := y[a] { P }

A very general approach – allows handling many data types

114

Array update rules (wp) example• Treat an array assignment y[a] := x as an

update to the array function y– y := y[ax] meaning y’=v. v=a ? x : y(v)

[array-update] { P[y[ax]/y] } y[a] := x { P }{x=y[i7](i)} y[i]:=7 {x=y(i)}

{x=7} y[i]:=7 {x=y(i)}

[array-load] { P[y(a)/x] } x := y[a] { P }{y(a)=7} x:=y[a] {x=7}

115

Array update rules (sp)

[array-updateF] { x=v y=g a=b } y[a] := x { y=g[bv] }

[array-loadF] { y=g a=b } x := y[a] { x=g(b) }

In both rulesv, g, and b are fresh

116

Array-max program

nums : arrayN : int // N stands for num’s length { N0 nums=orig_nums } x := 0res := nums[0]while x < N if nums[x] > res then res := nums[x] x := x + 11. { x=N }2. { m. (m0 m<N) nums(m)res }3. { m. m0 m<N nums(m)=res }4. { nums=orig_nums }

117

Array-max program

nums : arrayN : int // N stands for num’s length { N0 nums=orig_nums } x := 0res := nums[0]while x < N if nums[x] > res then res := nums[x] x := x + 1Post1: { x=N }Post2: { nums=orig_nums }Post3: { m. 0m<N nums(m)res }Post4: { m. 0m<N nums(m)=res }

118

Summary

• C programming language• P assertions• {P} C {Q} judgments• { P[a/x] } x := a { P } proof Rules– Soundness– Completeness

• {x = N} y:=factorial(x){ y = N!} proofs

119

Extensions to axiomatic semantics• Procedures• Total correctness assertions• Assertions for execution time– Exact time– Order of magnitude time

• Assertions for dynamic memory– Separation Logic

• Assertions for parallelism– Owicki-Gries– Concurrent Separation Logic– Rely-guarantee